\section{Line Graphs}

\problem{}
Given a graph $G$, we can construct a graph called the \par
\textit{line graph} of $G$ (\hspace{0.3ex}denoted $\mathcal{L}(G)$\hspace{0.3ex}) by doing the following: \par
\begin{itemize}
	\item Creating a node in $\mathcal{L}(G)$ for each edge in $G$
	\item Drawing a directed edge between every pair of nodes $a, b$ in $\mathcal{L}(G)$ \par
	if the corresponding edges in $G$ are adjacent. \par
	\note{That is, if edge $b$ in $G$ starts at the node at which $a$ ends.}
\end{itemize}

\problem{}
Draw the line graph for the graph below. \par
Have an instructor check your solution.

\begin{center}
	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[main] (a) at (0, 0) {$a$};
			\node[main] (b) at (2, 0) {$b$};
			\node[main] (c) at (4, 0) {$c$};
		\end{scope}

		\draw[->]
			(a) edge[bend left] node[label] {$0$} (b)
			(b) edge[bend left] node[label] {$1$} (a)
			(b) edge[bend left] node[label] {$2$} (c)
			(c) edge[bend left] node[label] {$3$} (b)
			(c) edge[loop right] node[label] {$4$} (c)
		;
	\end{tikzpicture}
\end{center}

\begin{solution}
	\begin{center}
		\begin{tikzpicture}
			\begin{scope}[layer = nodes]
				\node[main] (1) at (0, 0) {$1$};
				\node[main] (4) at (3.5, 1) {$4$};
				\node[main] (3) at (2, 0) {$3$};
				\node[main] (2) at (2, 2) {$2$};
				\node[main] (0) at (0, 2) {$0$};
			\end{scope}

			\draw[->]
				(0) edge[bend left] (1)
				(1) edge[bend left] (0)
				(0) edge (2)
				(2) edge (3)
				(2) edge[bend left] (4)
				(4) edge[bend left] (2)
				(3) edge (1)
				(4) edge (3)
				(4) edge[loop right] (4)
			;
		\end{tikzpicture}
	\end{center}
\end{solution}


\vfill

\definition{}
We say a graph $G$ is \textit{connected} if there is a path
between any two vertices of $G$.

\problem{}
Show that if $G$ is connected, $\mathcal{L}(G)$ is connected.

\begin{solution}
	Let $a, b$ and $x, y$ be nodes in a connected graph $G$ so that an edges $a \rightarrow b$ and
	and $x \rightarrow y$ exist. Since $G$ is connected, we can find a path from $b$ to $x$.
	The path $a$ to $y$ corresponds to a path in $\mathcal{L}(G)$ between $a \rightarrow b$ and $x \rightarrow y$.
\end{solution}

\vfill
\pagebreak


\definition{}
Consider $\mathcal{L}(G_n)$, where $G_n$ is the $n^\text{th}$ order De Bruijn graph. \par

\vspace{2mm}

We'll need to label the vertices of $\mathcal{L}(G_n)$. To do this, do the following:
\begin{itemize}
	\item Let $a$ and $b$ be nodes in $G_n$
	\item Let \texttt{x} be the first letter of $a$
	\item Let \texttt{y}, the last letter of $b$
	\item Let $\overline{\texttt{p}}$ be the prefix/suffix that $a$ and $b$ share. \par
	Note that $a = \texttt{x}\overline{\texttt{p}}$ and $b = \overline{\texttt{p}}\texttt{y}$,
\end{itemize}
Now, relabel the edge from $a$ to $b$ as $\texttt{x}\overline{\texttt{p}}\texttt{y}$. \par
Use these new labels to name nodes in $\mathcal{L}(G_n)$.

\problem{}
Construct $\mathcal{L}(G_2)$ and $\mathcal{L}(G_3)$. What do you notice? \par
\hint{
	What are $\mathcal{L}(G_2)$ and $\mathcal{L}(G_3)$? We've seen them before! \par
	You may need to re-label a few edges.
}

\begin{solution}
	After fixing edge labels, we find that
	$\mathcal{L}(G_2) \cong G_3$ and $\mathcal{L}(G_3) \cong G_4$
\end{solution}

\vfill
\pagebreak